\(\phi\)(x) = ||x - 1| - |x + 1||
\(=\begin{cases}
1-(x-1)-(-(x+1))&;x\leq-1\\
1-(x-1)-(x+1)&;-1\leq x\leq1\\
|x-1-(x+1)|&;x\geq1
\end{cases}\)
\(=\begin{cases}
2&;x\leq-1\\
2|x|&;-1\leq x\leq 1\\
2&;x\geq 1
\end{cases}\)
Now, f(x) = min\(\phi\)(f) ; -3 \(\leq t\leq x\)
\(=\begin{cases}
2&;-3\leq x\leq-1\,and\,x\geq 1\\
0&;-1\leq x\leq1
\end{cases}\)
\(\left(\because Min |t| = 0\\
for\,-1\leq t\leq x \leq 1\right)\)
\(\therefore\) f(x) is not differentiable at x = -1 and x = 1
(C) f(100) = 2 (\(\because 100>1\))
(D)
\(\int\limits_{-3}^{10}f(x)dx=\int\limits_{-3}^{-1}2dx+\int\limits_{-1}^10dx+\int\limits_1^{10}2dx\)
\(=2(x)^{-1}_{-3}+0+2(x)_1^{10}\)
= 2(-1-(-3)) + 2(10 - 1)
= 2(-1 + 3) + 2 x 9
= 4 + 18 = 22
